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[英]how can i load array to the function via pointers and use them for calculations in that function
[英]How can I use an array of function pointers?
我應該如何在 C 中使用函數指針數組?
我該如何初始化它們?
int sum(int a, int b);
int subtract(int a, int b);
int mul(int a, int b);
int div(int a, int b);
int (*p[4]) (int x, int y);
int main(void)
{
int result;
int i, j, op;
p[0] = sum; /* address of sum() */
p[1] = subtract; /* address of subtract() */
p[2] = mul; /* address of mul() */
p[3] = div; /* address of div() */
[...]
要調用這些函數指針之一:
result = (*p[op]) (i, j); // op being the index of one of the four functions
以上答案可能對您有所幫助,但您可能還想知道如何使用函數指針數組。
void fun1()
{
}
void fun2()
{
}
void fun3()
{
}
void (*func_ptr[3])() = {fun1, fun2, fun3};
main()
{
int option;
printf("\nEnter function number you want");
printf("\nYou should not enter other than 0 , 1, 2"); /* because we have only 3 functions */
scanf("%d",&option);
if((option>=0)&&(option<=2))
{
(*func_ptr[option])();
}
return 0;
}
您只能將具有相同返回類型和相同參數類型且沒有參數的函數地址分配給單個函數指針數組。
如果上述所有函數都具有相同數量的相同類型的參數,您還可以傳遞如下參數。
(*func_ptr[option])(argu1);
注意:在數組中,函數指針的編號將從 0 開始,與一般數組相同。 因此,在上面的例子中fun1
可以稱為如果選項= 0, fun2
可以稱為如果選項= 1和fun3
的話可以稱為選項= 2。
使用方法如下:
#ifndef NEW_FUN_H_
#define NEW_FUN_H_
#include <stdio.h>
typedef int speed;
speed fun(int x);
enum fp {
f1, f2, f3, f4, f5
};
void F1();
void F2();
void F3();
void F4();
void F5();
#endif
#include "New_Fun.h"
speed fun(int x)
{
int Vel;
Vel = x;
return Vel;
}
void F1()
{
printf("From F1\n");
}
void F2()
{
printf("From F2\n");
}
void F3()
{
printf("From F3\n");
}
void F4()
{
printf("From F4\n");
}
void F5()
{
printf("From F5\n");
}
#include <stdio.h>
#include "New_Fun.h"
int main()
{
int (*F_P)(int y);
void (*F_A[5])() = { F1, F2, F3, F4, F5 }; // if it is int the pointer incompatible is bound to happen
int xyz, i;
printf("Hello Function Pointer!\n");
F_P = fun;
xyz = F_P(5);
printf("The Value is %d\n", xyz);
//(*F_A[5]) = { F1, F2, F3, F4, F5 };
for (i = 0; i < 5; i++)
{
F_A[i]();
}
printf("\n\n");
F_A[f1]();
F_A[f2]();
F_A[f3]();
F_A[f4]();
return 0;
}
我希望這有助於理解Function Pointer.
這個“答案”更像是 VonC 答案的附錄; 請注意,可以通過 typedef 簡化語法,並且可以使用聚合初始化:
typedef int FUNC(int, int);
FUNC sum, subtract, mul, div;
FUNC *p[4] = { sum, subtract, mul, div };
int main(void)
{
int result;
int i = 2, j = 3, op = 2; // 2: mul
result = p[op](i, j); // = 6
}
// maybe even in another file
int sum(int a, int b) { return a+b; }
int subtract(int a, int b) { return a-b; }
int mul(int a, int b) { return a*b; }
int div(int a, int b) { return a/b; }
哦,有很多例子。 只需看看 glib 或 gtk 中的任何內容。 您可以在那里一直看到函數指針的工作。
例如,這里是 gtk_button 東西的初始化。
static void
gtk_button_class_init (GtkButtonClass *klass)
{
GObjectClass *gobject_class;
GtkObjectClass *object_class;
GtkWidgetClass *widget_class;
GtkContainerClass *container_class;
gobject_class = G_OBJECT_CLASS (klass);
object_class = (GtkObjectClass*) klass;
widget_class = (GtkWidgetClass*) klass;
container_class = (GtkContainerClass*) klass;
gobject_class->constructor = gtk_button_constructor;
gobject_class->set_property = gtk_button_set_property;
gobject_class->get_property = gtk_button_get_property;
在 gtkobject.h 中,您可以找到以下聲明:
struct _GtkObjectClass
{
GInitiallyUnownedClass parent_class;
/* Non overridable class methods to set and get per class arguments */
void (*set_arg) (GtkObject *object,
GtkArg *arg,
guint arg_id);
void (*get_arg) (GtkObject *object,
GtkArg *arg,
guint arg_id);
/* Default signal handler for the ::destroy signal, which is
* invoked to request that references to the widget be dropped.
* If an object class overrides destroy() in order to perform class
* specific destruction then it must still invoke its superclass'
* implementation of the method after it is finished with its
* own cleanup. (See gtk_widget_real_destroy() for an example of
* how to do this).
*/
void (*destroy) (GtkObject *object);
};
(*set_arg) 東西是一個指向函數的指針,這可以例如在某些派生類中分配另一個實現。
經常看到這樣的東西
struct function_table {
char *name;
void (*some_fun)(int arg1, double arg2);
};
void function1(int arg1, double arg2)....
struct function_table my_table [] = {
{"function1", function1},
...
因此,您可以按名稱進入表並調用“關聯”函數。
或者,您可能使用一個哈希表,在其中放置函數並“按名稱”調用它。
問候
弗里德里希
可以像這樣使用它:
//! Define:
#define F_NUM 3
int (*pFunctions[F_NUM])(void * arg);
//! Initialise:
int someFunction(void * arg) {
int a= *((int*)arg);
return a*a;
}
pFunctions[0]= someFunction;
//! Use:
int someMethod(int idx, void * arg, int * result) {
int done= 0;
if (idx < F_NUM && pFunctions[idx] != NULL) {
*result= pFunctions[idx](arg);
done= 1;
}
return done;
}
int x= 2;
int z= 0;
someMethod(0, (void*)&x, &z);
assert(z == 4);
這是一個更簡單的示例,說明如何執行此操作:
跳轉表.c
int func1(int arg) { return arg + 1; }
int func2(int arg) { return arg + 2; }
int func3(int arg) { return arg + 3; }
int func4(int arg) { return arg + 4; }
int func5(int arg) { return arg + 5; }
int func6(int arg) { return arg + 6; }
int func7(int arg) { return arg + 7; }
int func8(int arg) { return arg + 8; }
int func9(int arg) { return arg + 9; }
int func10(int arg) { return arg + 10; }
int (*jump_table[10])(int) = { func1, func2, func3, func4, func5,
func6, func7, func8, func9, func10 };
int main(void) {
int index = 2;
int argument = 42;
int result = (*jump_table[index])(argument);
// result is 45
}
存儲在數組中的所有函數必須具有相同的簽名。 這只是意味着它們必須返回相同的類型(例如int
)並具有相同的參數(在上面的示例中為單個int
)。
在 C++ 中,您可以對靜態類方法(但不是實例方法)執行相同的操作。 例如,您可以在上面的數組中使用MyClass::myStaticMethod
但不能使用MyClass::myInstanceMethod
或instance.myInstanceMethod
:
class MyClass {
public:
static int myStaticMethod(int foo) { return foo + 17; }
int myInstanceMethod(int bar) { return bar + 17; }
}
MyClass instance;
這應該是上述響應的簡短且簡單的復制和粘貼代碼示例。 希望這會有所幫助。
#include <iostream>
using namespace std;
#define DBG_PRINT(x) do { std::printf("Line:%-4d" " %15s = %-10d\n", __LINE__, #x, x); } while(0);
void F0(){ printf("Print F%d\n", 0); }
void F1(){ printf("Print F%d\n", 1); }
void F2(){ printf("Print F%d\n", 2); }
void F3(){ printf("Print F%d\n", 3); }
void F4(){ printf("Print F%d\n", 4); }
void (*fArrVoid[N_FUNC])() = {F0, F1, F2, F3, F4};
int Sum(int a, int b){ return(a+b); }
int Sub(int a, int b){ return(a-b); }
int Mul(int a, int b){ return(a*b); }
int Div(int a, int b){ return(a/b); }
int (*fArrArgs[4])(int a, int b) = {Sum, Sub, Mul, Div};
int main(){
for(int i = 0; i < 5; i++) (*fArrVoid[i])();
printf("\n");
DBG_PRINT((*fArrArgs[0])(3,2))
DBG_PRINT((*fArrArgs[1])(3,2))
DBG_PRINT((*fArrArgs[2])(3,2))
DBG_PRINT((*fArrArgs[3])(3,2))
return(0);
}
最簡單的解決方案是給出你想要的最終向量的地址,並在函數內部修改它。
void calculation(double result[] ){ //do the calculation on result
result[0] = 10+5;
result[1] = 10 +6;
.....
}
int main(){
double result[10] = {0}; //this is the vector of the results
calculation(result); //this will modify result
}
這個問題已經用很好的例子回答了。 唯一可能缺少的示例是函數返回指針的示例。 我用這個寫了另一個例子,並添加了很多評論,以防有人發現它有幫助:
#include <stdio.h>
char * func1(char *a) {
*a = 'b';
return a;
}
char * func2(char *a) {
*a = 'c';
return a;
}
int main() {
char a = 'a';
/* declare array of function pointers
* the function pointer types are char * name(char *)
* A pointer to this type of function would be just
* put * before name, and parenthesis around *name:
* char * (*name)(char *)
* An array of these pointers is the same with [x]
*/
char * (*functions[2])(char *) = {func1, func2};
printf("%c, ", a);
/* the functions return a pointer, so I need to deference pointer
* Thats why the * in front of the parenthesis (in case it confused you)
*/
printf("%c, ", *(*functions[0])(&a));
printf("%c\n", *(*functions[1])(&a));
a = 'a';
/* creating 'name' for a function pointer type
* funcp is equivalent to type char *(*funcname)(char *)
*/
typedef char *(*funcp)(char *);
/* Now the declaration of the array of function pointers
* becomes easier
*/
funcp functions2[2] = {func1, func2};
printf("%c, ", a);
printf("%c, ", *(*functions2[0])(&a));
printf("%c\n", *(*functions2[1])(&a));
return 0;
}
這個帶有函數指針的多維數組的簡單示例”:
void one( int a, int b){ printf(" \n[ ONE ] a = %d b = %d",a,b);}
void two( int a, int b){ printf(" \n[ TWO ] a = %d b = %d",a,b);}
void three( int a, int b){ printf("\n [ THREE ] a = %d b = %d",a,b);}
void four( int a, int b){ printf(" \n[ FOUR ] a = %d b = %d",a,b);}
void five( int a, int b){ printf(" \n [ FIVE ] a = %d b = %d",a,b);}
void(*p[2][2])(int,int) ;
int main()
{
int i,j;
printf("multidimensional array with function pointers\n");
p[0][0] = one; p[0][1] = two; p[1][0] = three; p[1][1] = four;
for ( i = 1 ; i >=0; i--)
for ( j = 0 ; j <2; j++)
(*p[i][j])( (i, i*j);
return 0;
}
#include <iostream>
using namespace std;
int sum (int , int);
int prod (int , int);
int main()
{
int (*p[2])(int , int ) = {sum,prod};
cout << (*p[0])(2,3) << endl;
cout << (*p[1])(2,3) << endl;
}
int sum (int a , int b)
{
return a+b;
}
int prod (int a, int b)
{
return a*b;
}
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